JEE Main 2026 Apr 4 Shift 2, Mathematics Q21: Introduction To Probability

JEE Main2026Apr 4, Shift 2Mathematics

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{a}{b}$ , where $a, b \in N$ and $g cd (a, b) = 1$ , then $a + b$ is equal to ______.

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Solution

Total ways to select 3 dates: $(3 31) = \frac{31 \cdot 30 \cdot 29}{6} = 4495$ .

For three numbers $a, b, c$ in A.P. (increasing): $a + c = 2 b$ , so $a$ and $c$ must have the same parity, and $b$ is determined uniquely.

Count favourable: choose any two numbers with the same parity from ${1, 2, \dots, 31}$ . Odds (1, 3, ..., 31): 16 numbers; evens: 15 numbers.

Favourable $= (2 16) + (2 15) = 120 + 105 = 225$ .

Probability $= \frac{225}{4495} = \frac{45}{899}$ (dividing both by 5; $g cd (45, 899) = 1$ since $899 = 29 \times 31$ ).

$a + b = 45 + 899 = 944$ .

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JEE Main 2026 Apr 4 Shift 2, Mathematics Q21: Introduction To Probability

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